Each morning John eats some eggs. On any given morning, the number of eggs he eats is equally likely to be 1, 2, 3, 4, 5, or 6 independent of what he has done in the past. Let X be the number of eggs that John eats in 10 days. Find the mean and variance of X.
First, I have E[X] = sum of E[Xi] from 1 to 10 = [(1+2+3+4+5+6)/6]*10 = 35. Then, I get var(X) = sum of var(Xi) from 1 to 10 =[(1^2+2^2+3^2+4^2+5^2+6^2)/6]*10^2 -(35)^2 = 291.667.
However, I found another answer by using this formula
Var(Xi) = ((6-1)(6-1+2)/12) = 2.9167, and Var(X) = sum of var(Xi) from 1 to 10 = 29.167.
The two answer are off from each other by 10. Why?
The 1st solution is wrong: $$E[X_i]=(1+2+3+4+5+6)/6=3.5$$ $$Var(X_i)=E[X_i^2]-(E[X_i])^2=(1^2+2^2+3^2+4^2+5^2+6^2)/6-3.5^2\approx 2.9167$$